The surface area of a cube is increasing at a rate of $15$ square meters per hour. At a certain instant, the surface area is $24$ square meters. What is the rate of change of the volume of the cube at that instant (in cubic meters per hour)? Choose 1 answer: Choose 1 answer: (Choice A) A $(\sqrt{15})^{^3}$ (Choice B) B $\dfrac{15}{2}$ (Choice C) C $\dfrac{5}{8}$ (Choice D) D $8$
Setting up the math Let... $s(t)$ denote the cube's side at time $t$, $V(t)$ denote the cube's volume at time $t$, and $S(t)$ denote the cube's surface area at time $t$. We are given that $S'(t)=15$, We are also given that $S(t_0)=24$ for a specific time $t_0$. We want to find $V'(t_0)$. Relating the measures $S(t)$ and $s(t)$ relate to each other through the formula for the surface area of a cube: $S(t)=6[s(t)]^2$ We can differentiate both sides to find an expression for $S'(t)$ : $S'(t)=12s(t)s'(t)$ $V(t)$ and $s(t)$ relate to each other through the formula for the volume of a cube: $V(t)=[s(t)]^3$ We can differentiate both sides to find an expression for $V'(t)$ : $V'(t)=3[s(t)]^2s'(t)$ Using the information to solve Let's plug ${S(t_0)}={24}$ into the expression for $S(t_0)$ : $\begin{aligned} {S(t_0)}&=6[s(t_0)]^2 \\\\ {24}&=6[s(t_0)]^2 \\\\ 4&=[s(t_0)]^2 \\\\ {2}&={s(t_0)} \end{aligned}$ Let's plug ${S'(t_0)}={15}$ and ${s(t_0)}={2}$ into the expression for $S'(t_0)$ : $\begin{aligned} {S'(t_0)}&=12{s(t_0)}s'(t_0) \\\\ {15}&=12({2})s'(t_0) \\\\ C{\dfrac{5}{8}}&=C{s'(t_0)} \end{aligned}$ Now let's plug ${s(t_0)}={2}$ and $C{s'(t_0)}=C{\dfrac{5}{8}}$ into the expression for $V'(t_0)$ : $\begin{aligned} V'(t_0)&=3[{s(t_0)}]^2C{s'(t_0)} \\\\ &=3\left({2}\right)^2\left(C{\dfrac{5}{8}}\right) \\\\ &=\dfrac{15}{2} \end{aligned}$ In conclusion, the rate of change of the volume of the cube at that instant is $\dfrac{15}{2}$ cubic meters per hour. Since the rate of change is positive, we know that the volume is increasing.